Question

A cube of aluminium each side 4cm is subjected to a tangential or shear force . the top face of the cube is sheared through 0.012cm with respect to the bottom face ?
FIND a. shearing strain
b. shearing stress
c. shearing force
n=2x10^​11 dyes/ cm²

Answer 1

$$\begin{gathered} \mathrm{Given}, \\ \mathrm{Length}\mathrm{of}\mathrm{each}\mathrm{side},\mathrm{L}=4\mathrm{cm} \\ \mathrm{Lateral}\mathrm{displacement},\mathrm{x}=0.012\mathrm{cm} \\ \mathrm{\eta }=2\times {10}^{11}\mathrm{dyne}/{\mathrm{cm}}^2 \\ \left(\mathrm{a}\right)\mathrm{Shearing}\mathrm{strain}\mathrm{\theta }=\frac{\mathrm{x}}{\mathrm{L}}=\frac{0.012}{4}=0.003 \\ \left(\mathrm{b}\right)\mathrm{Using}\mathrm{relation}, \\ \mathrm{\eta }=\frac{\mathrm{F}}{\mathrm{\theta a}} \\ \mathrm{or}\mathrm{Shearing}\mathrm{stress}=\frac{\mathrm{F}}{\mathrm{a}}=\mathrm{\eta \theta } \\ =2\times {10}^{11}\times 3\times {10}^{-3} \\ =6\times {10}^8\mathrm{dyne}/{\mathrm{cm}}^2 \\ \left(\mathrm{c}\right)\mathrm{Shearing}\mathrm{force},\mathrm{F}=\mathrm{Shearing}\mathrm{stress}\times \mathrm{area} \\ =6\times {10}^8\times {\mathrm{L}}^2 \\ =6\times {10}^8\times 4^2 \\ =9.6\times {10}^9\mathrm{dyne} \end{gathered}$$